When you select a random person, the probability that this person will go to Ches
Tahoe restaurant is 75% (Hint: Review Binomial Probability Distribution).
a. You randomly select 8 people. What is the probability that exactly 3 people
from that group will go to Ches Tahoe restaurant? (You must show the
steps to get the full credit) (15 points)
b. You randomly select 50 people. Use the normal approximation to the binomial
distribution find the probability that at least 35 people will go to Ches Tahoe
restaurant (Hint: mean = np: stand dev = sqrt(npq), where p + y = ?). (No
steps necessary - just write the answer) (5 points)

Respuesta :

Answer: a= 0.0231

Step-by-step explanation:

n=8

p=0.75

q=1 - 0.75 = 0.25

(p=x)

p(5) = 8C5 (0.25)^5 (0.75)^8-5

p(5)= 0.0231

Or 2.31%

The probability that exactly 3 people  from that group will go to Ches Tahoe restaurant is 2.11%

The probability that at least 35 people will go to Ches Tahoe  the restaurant is -1.88.

Given that,

When you select a random person, the probability that this person will go to Ches Tahoe restaurant is 75%,

We have to determine,

You randomly select 8 people. What is the probability that exactly 3 people

from that group will go to Ches Tahoe restaurant.

You randomly select 50 people. Use the normal approximation to the binomial distribution to find the probability that at least 35 people will go to Ches Tahoe restaurant.

According to the question,

When you select a random person, the probability that this person will go to Ches Tahoe restaurant is 75%,

You randomly select 8 people.

Then,

The probability that exactly 3 people

from that group will go to Ches Tahoe restaurant is,

[tex]P(x) = ^nC_x \times q^x \times p^{n-x}[/tex]

Where n = 8, x = 5, q = 1 - 0.75 = 0.25, and p = 0.75

Substitute all the values in the expression,

[tex]P(5) = ^8C_5 \times (0.25)^5 \times (0.75)^{8-5}\\\\P(5) = \dfrac{8!}{(8-5)!. 5!} \times 0.00097 \times (0.75)^3\\\\P(5) = \dfrac{8 \times 7 \times 6}{3 \times 2} \times 0.00097 \times 0.42\\\\P(5) = 0.2118\\\\P(5) = 2.11 \ Percent[/tex]

The probability that exactly 3 people from that group will go to Ches Tahoe restaurant is 2.11%.

You randomly select 50 people. Use the normal approximation to the binomial  distribution

Then, the probability that at least 35 people will go to Ches Tahoe  restaurant is,

Mean for that at least 35 people will go to Ches Tahoe  restaurant is,

[tex]\mu = n \times p \\\\\mu = 8 \times 0.75\\\\\mu = 6[/tex]

And standard deviation that at least 35 people will go to Ches Tahoe  restaurant is,

[tex]\sigma = \sqrt{n \times p \times q} \\\\\sigma = \sqrt{0.75 \times 0.25 \times 8}\\\\\sigma = \sqrt{1.5}\\\\\sigma = 0.53[/tex]

Normal approximation to the binomial distribution is,

[tex]z = \dfrac{x-\mu}{\sigma}\\\\z = \dfrac{5-6}{0.53}\\\\z = \dfraac{-1}{0.53}\\\\z = -1.88[/tex]

The probability that at least 35 people will go to Ches Tahoe  the restaurant is -1.88.

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